Convergence of $f_n(x) = x+ (1/n) $ for $x \in \mathbb R$ Let $f_n(x) = x+ (1/n) $ for $x \in \mathbb  R$; 
I must prove that $f_n$ converges to $F(x) = x$ uniformly on $\mathbb R$, and that $f_n^2$ does not converge uniformly. How can this be proven? 
 A: Note that $|f_n(x)-f(x)|=\frac{1}{n}$ for all $n$ and $x$. Let $\epsilon>0$. Since $\frac{1}{n}\to0$, there is $N>0$ such that $\frac{1}{n}<\epsilon$ for all $n\ge N$. So, if $n\ge N$ we have $|f_n(x)-f(x)|=\frac{1}{n}<\epsilon$ independentely of $x$. Then $\{f_n\}$ is uniformly convergent to $f$.
For the second part, if $g_n=f_n^2$ would be uniformly convergent, then it should be pointwise convergent. But $\lim_{n\to\infty}g_n(x)=\lim_{n\to\infty}x^2+\frac{2x}{n}+\frac{1}{n^2}=x^2$. Thus, if $\{g_n\}$ is uniformly convergent, then it converges to $g(x)=x^2$.
Suppose there is uniform convergence. That is, there is some $N>0$ such that $|g_n(x)-g(x)|<1$ for all $n\ge N$, which not depends on $x$. That is, for all $x$ we have $|\frac{2x}{n}+\frac{1}{n^2}|<1$ for $n\ge N$. In particular $|\frac{2x}{N}+\frac{1}{N^2}|<1$ FOR ALL $x$. But since $\lim_{x\to\infty}\frac{2x}{N}+\frac{1}{N^2}=\infty$, then there is $x_0>0$ such that $|\frac{2x_0}{N}+\frac{1}{N^2}|\ge 1$. This contradiction shows that $\{g_n\}$ is not uniformly convergent.
A: For the first part (the uniform convergence part), just use the definition.
Pick an arbitrary $\epsilon > 0$. Show that there exists $N$ (which may depend on $\epsilon$, but not on $x$), such that when $n > N$, we have: $|f_n(x) - F(x)| < \epsilon$ (for all $x$). Obviously here you can use $N\geq \frac{1}{\epsilon}$.     
The second statement is not true. The function sequence $f_n^2(x)$ converges to $G(x) = x^2$. Maybe you meant that it does not converge uniformly to $x^2$. Again, use the definition of uniform convergence.   
